IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

agnit_thebest

The domain of definition of the function y(x) given by the equation a^x+a^y=a(a>1) is
a) 0<x<=1
b) 0<=x<1
c) -infinity<x<1
d) -infinity<x<=0

ramyadiamond

I think its c.

a^y = a-a^x

as a>1

hence it is safer to take log on both sides

y.loga=log(a-a^x)

hence, a-a^x >0

a^x<a

x.loga<loga

x<1

Hence, i think it should be c.

Do correct me if i'm wrong.

agnit_thebest

Sorry the answer given is b.

pottermania1990

try substituiting a as 2,b as -2

condition c and are not satisfied

by substit. we can eleminate a c and d

ans . is b

i noe my reply is not up 2 the mark

asmit

a^x +a^y=a or a^(x-1)+a^(y-1)=1 {a not equal to 0}

so since a>1 therefore to make each less than 1, x-1<0

or x<1 The other end of the domain has to be -infinity because y can be adjusted likewise. It should be (c),

For ex. put x= -2 you get with a=2

2^(-3)+2^(y-1)=1 or 2^(y-1)=7/8 therefore (y-1)= -.19 or y=.8

which satisfies the condition

joyfrancis

Dont you think it should be a^x+a^y(x)=a

Since y is a function.

Like if we have any function f(x)... a^fhas no meaning.

Please See.

asmit

Why? the question is perfectly alright.We don't need to worry about whether y is a func. of x..Infact the above question is valid for n variables that is x,y,z,w,p,q,......

a^x+a^y+a^z+a^w+a^p.......=a

aansh_c

1stly hw did u gt d power up dere temme also

a(power of x) + a(power of y)=a?

a(power of y)=a-a(power of x)

take log 2 d base a on both sides

y=log(a-a(power of x)) 2 d base of a

now a-a(power of x) should be>0

i.e. a(power of of x)<a

nw since a is >1

d inequality sign doesnt chng & it bcoms

x<1

savvej

ans is b.rayamond is right n also o satisfies the function and the domain cant be negative.hence.

ramyadiamond

as far as i can see, there's no problem in taking x negative. so hy cant it be (c)??

savvej

@ raymond give example

asmit

Savvej I have given one example

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